SOLUTION: NASA launches a rocker at T=0 seconds. It's height, in meters above sea level, as a function of time is given by H(T)=-4.9T^2 + 229T +185. Assuming that the rocket will splash

Question 1136257: NASA launches a rocker at T=0 seconds. It's height, in meters above sea level, as a function of time is given by H(T)=-4.9T^2 + 229T +185.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after ___ seconds
How high above sea-level does the rocket get at its peak?
The rocket peaks at ___ meters above sea-level

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
ASA launches a rocker at T=0 seconds. It's height, in meters above sea level, as a function of time is given by H(T)=-4.9T^2 + 229T +185.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
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it's = it is.
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I would solve the problem, but these entries irritate me.
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The rocket splashes down after ___ seconds
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How high above sea-level does the rocket get at its peak?
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The rocket peaks at ___ meters above sea-level
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this is not useful.
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H(T)=-4.9T^2 + 229T +185
At impact, h(t) = 0
-4.9T^2 + 229T +185 = 0
Solve for t

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=56067 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.794358402485433, 47.5290522800364. Here's your graph:

Ignore the negative value.
t = ~ 47.529 seconds
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The peak (max height) is the vertex of the parabola, at t = -b/2a
t = -229/-9.8 = ~ 23.367 seconds
h(23.367) =~ 2675.5 meters
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PS - this is not a rocket, it's a projectile.
Rockets accelerate upward.
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